Memory overlap
length of the array of n m data to the front
void the Move (ARR int *, int n, int m)
{
IF (m <0 || = m> = n)
{
return;
}
int * BRR = (int ) the malloc (m the sizeof (int)); // int BRR [m];
int i;
for(i=0;i<m;i++) //将后面的m个数据另外保存
{
brr[i] = arr[n-m+i];
}
for(i=n-m-1;i>=0;i--)//将数据往后移
{
arr[i+m] = arr[i];
}
//将brr数据移到arr中
for(i=0;i<m;i++)
{
arr[i] = brr[i];
}
free(brr);
}
int main()
{
int arr[10] = {1,2,3,4,5,6,7,8,9,10};
Move(arr,sizeof(arr)/sizeof(arr[0]),3);
Show(arr,sizeof(arr)/sizeof(arr[0]));
return 0;
}
Memory aligned
address a variety of data to be stored can be divisible by the number of bytes of data per se